From a lot containing 10 defective and 90 non | vedclass

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(D) Total bulbs = $100$. Number of defective bulbs = $10$. Number of non-defective bulbs = $90$.Probability of selecting a defective bulb $p = \frac{1

Vedclass · Accepted Answer

$\frac{73}{10^{8}}$

Vedclass · Answer

(D) Total bulbs = $100$. Number of defective bulbs = $10$. Number of non-defective bulbs = $90$.Probability of selecting a defective bulb $p = \frac{10}{100} = \frac{1}{10}$.Probability of selecting a non-defective bulb $q = \frac{90}{100} = \frac{9}{10}$.Since bulbs are selected with replacement,this follows a Binomial Distribution $B(n, p)$ with $n = 8$ and $p = 0.1$.The probability of getting at least $7$ defective bulbs is $P(X \ge 7) = P(X = 7) + P(X = 8)$.$P(X = 7) = \binom{8}{7} 	imes (0.1)^7 	imes (0.9)^1 = 8 	imes \frac{1}{10^7} 	imes \frac{9}{10} = \frac{72}{10^8}$.$P(X = 8) = \binom{8}{8} 	imes (0.1)^8 	imes (0.9)^0 = 1 	imes \frac{1}{10^8} 	imes 1 = \frac{1}{10^8}$.Total probability = $\frac{72}{10^8} + \frac{1}{10^8} = \frac{73}{10^8}$.

From a lot containing $10$ defective and $90$ non-defective bulbs,$8$ bulbs are selected one by one with replacement. Then the probability of getting at least $7$ defective bulbs is:

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